Question

If ${\tan }^{-1}\frac{x-1}{x-2}+{\tan }^{-1}\frac{x+1}{x+2}=\frac{\pi }{4}$, then find the value of $x$.

Answer 1

${\tan }^{-1}\frac{x-1}{x-2}+{\tan }^{-1}\frac{x+1}{x+2}=\frac{\pi }{4}$ ...Given

$\Rightarrow {\tan }^{-1}\left[\frac{\frac{x-1}{x-2}+\frac{x+1}{x+2}}{1-\left(\frac{x-1}{x-2}\right)\left(\frac{x+1}{x+2}\right)}\right]=\frac{\pi }{4}$

$\Rightarrow {\tan }^{-1}\left[\frac{(x-1)(x+2)+(x+1)(x-2)}{(x+2)(x-2)-(x-1)(x+1)}\right]=\frac{\pi }{4}$

$\Rightarrow {\tan }^{-1}\left[\frac{x^2+x-2+x^2-x-2}{x^2-4-x^2+1}\right]=\frac{\pi }{4}$

$\Rightarrow {\tan }^{-1}\left[\frac{2x^2-4}{-3}\right]=\frac{\pi }{4}$

$\Rightarrow \tan \left[{\tan }^{-1}\frac{4-{2x}^2}{3}\right]=\tan \frac{\pi }{4}$

$\Rightarrow \frac{4-{2x}^2}{3}=1$

$\Rightarrow 4-2x^2=3$

$\Rightarrow 2x^2=4-3=1$

$\Rightarrow x=\pm \frac{1}{\sqrt{2}}$